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Chapter 5: Problem 612

A uniform disc of mass \(500 \mathrm{~kg}\) and radius \(2 \mathrm{~m}\) is rotating at the rate of \(600 \mathrm{r}\).p.m. what is the torque required to rotate the disc in the opposite direction with the same angular speed in a time of \(100 \mathrm{sec}\) ? \(\\{\mathrm{A}\\} 600 \pi \mathrm{Nm}\) \\{B \(\\} 500 \pi \mathrm{Nm}\) \\{C \(\\} .400 \pi \mathrm{Nm}\) \(\\{\mathrm{D}\\} 300 \pi \mathrm{Nm}\)

Short Answer

Expert verified
The torque required to rotate the disc in the opposite direction with the same angular speed in a time of 100 sec is approximately \(400\pi Nm\).

Step by step solution

01

Calculate the moment of inertia of the disc

The formula for the moment of inertia of a uniform disc is given by: \[I = \frac{1}{2}MR^2\] where M is the mass of the disc, R is its radius, and I is the moment of inertia. Given the mass M = 500 kg and radius R = 2 m, we can plug these values into the formula: \[I = \frac{1}{2} \times 500 \times (2)^2 = 1000\,kg.m^2\]
02

Convert angular speed from RPM to rad/s

Currently, the disc is rotating at 600 r.p.m., which should be converted to rad/s in order to find the angular acceleration later on. 1 revolution equals \(2\pi\) radians, and 1 minute equals 60 seconds. Therefore, \[\omega_1 = 600 \frac{rev}{min} \times \frac{2\pi\, rad}{1\, rev} \times \frac{1\, min}{60\, s} = 62.83\, rad/s\] where \(\omega_1\) is the initial angular speed.
03

Calculate the final angular speed in the opposite direction

As the final angular speed has the same magnitude but is in the opposite direction, we can represent it as the negative of the initial angular speed. \[\omega_2 = -\omega_1 = -62.83\, rad/s\]
04

Calculate the angular acceleration

We have the initial and final angular speeds as well as the time taken for the change. Using the equation \(\alpha = \frac{\omega_2 - \omega_1}{t}\), we can find the angular acceleration. Given the time t = 100s, we can calculate the angular acceleration as: \[\alpha = \frac{-62.83 - 62.83}{100} = -1.2566\, rad/s^2\]
05

Calculate the torque required

To calculate the torque required, we can use the formula: \[\tau = I\alpha\] Plugging in the moment of inertia I = 1000 kg.m虏 and angular acceleration \(\alpha = -1.2566\, rad/s^2\), we get: \[\tau = 1000 \times (-1.2566) = -1256.6\, Nm\] Since torque is a vector quantity, we need to consider its magnitude only. So, the required torque magnitude is: \[\tau = 1256.6\, Nm\] Now, let us see which option matches our result: \(600\pi Nm \approx 1884.96 Nm\) \\ \(500\pi Nm \approx 1570.80 Nm\) \\ \(400\pi Nm \approx 1256.64 Nm\) \\ \(300\pi Nm \approx 942.48 Nm\) As our result is very close to 1256.64 Nm, we choose the closest option C: \[ \tau = 400\pi Nm\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a concept crucial to understanding rotational motion. It's akin to mass in linear motion, but instead applies to how mass is distributed relative to an axis of rotation. For a disc, the moment of inertia is determined using the formula:\[I = \frac{1}{2} MR^2\]Where:
  • \( M \) is the mass of the disc.
  • \( R \) is the radius.
  • \( I \) is the moment of inertia.
Plugging in the known values from the exercise like \( M = 500 \, kg \) and \( R = 2 \, m \), we compute \( I \) as \( 1000 \, kg \cdot m^2 \). This means the disc's inertia depends on both its mass and how the mass is spread out relative to the center.
Angular Acceleration
Once you have converted the angular speeds to radian units, you can calculate angular acceleration. It represents how quickly the rotational speed of the disc changes. Defined mathematically as the change in angular speed over time:\[\alpha = \frac{\omega_2 - \omega_1}{t}\]Where:
  • \( \omega_1 \) is the initial angular speed.
  • \( \omega_2 \) is the final angular speed.
  • \( t \) is the time duration over which this change occurs.
For the exercise, the disc's initial speed was \( 62.83 \, rad/s \), and it needed to reach \(-62.83 \, rad/s\) (opposite direction) over 100 seconds. Thus, we found \( \alpha = -1.2566 \, rad/s^2 \). The negative sign indicates a deceleration.
Radial Motion
Radial motion in the context of rotational dynamics often refers to the path along which the masses in the rotating object follow. In a uniformly rotating disc, all points travel in circular paths around the axis of rotation. The radial position does not change as you measure it from the center. However, each point on the disc has a different linear speed. The points further from the axis move faster in terms of linear speed even though they share the same angular speed.
Conversion of Units
Before we can fully solve problems on rotational dynamics, it's essential to convert units into a consistent system. In this exercise, the angular speed was initially given in revolutions per minute (RPM), which needed to be converted to radians per second.The conversion process is as follows:
  • 1 revolution = \( 2\pi \) radians.
  • 1 minute = 60 seconds.
Thus, to convert the angular speed from RPM to rad/s, use the formula:\[\omega = RPM \times \frac{2\pi\, rad}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ sec}}\]By applying this to the exercise, \( 600 \, RPM \) was converted to \( 62.83 \, rad/s \), allowing subsequent calculations for acceleration and torque.

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Most popular questions from this chapter

A wheel of mass \(10 \mathrm{~kg}\) has a moment of inertia of \(160 \mathrm{~kg} \mathrm{~m}\) radius of gyration will be \(\begin{array}{llll}\\{\mathrm{A}\\} 10 & \\{\mathrm{~B}\\} 8 & \\{\mathrm{C}\\} 6 & \\{\mathrm{D}\\} 4\end{array}\)

A car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

Two spheres each of mass \(\mathrm{M}\) and radius \(\mathrm{R} / 2\) are connected with a mass less rod of length \(2 \mathrm{R}\) as shown in figure. What will be moment of inertia of the system about an axis passing through centre of one of the spheres and perpendicular to the rod? \(\\{\mathrm{A}\\}(21 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(2 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(5 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(5 / 21) \mathrm{MR}^{2}\)

A player caught a cricket ball of mass \(150 \mathrm{gm}\) moving at a rate of \(20 \mathrm{~m} / \mathrm{s}\) If the catching process is Completed in \(0.1\) sec the force of the flow exerted by the ball on the hand of the player ..... N \(\\{\mathrm{A}\\} 3\) \(\\{B\\} 30\) \(\\{\mathrm{C}\\} 150\) \(\\{\mathrm{D}\\} 300\)

A solid sphere and a solid cylinder having same mass and radius roll down the same incline the ratio of their acceleration will be.... \(\\{\mathrm{A}\\} 15: 14\) \(\\{\mathrm{B}\\} 14: 15\) \(\\{\mathrm{C}\\} 5: 3\) \(\\{\mathrm{D}\\} 3: 5\)
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